Question

Use f’( x ) = lim With h ---> 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²

Answer 1

The derivative of given function is -2x.

Explanation:

The first principle of differentiation is

`f'(x)=lim_(h\rightarrow 0)(f(x+h)-f(x))/(h)`

The given function is

`f(x)=5-x^2`

`f'(x)=lim_(h\rightarrow 0)(5-(x+h)^2-(5-h^2)/(h)`

`f'(x)=lim_(h\rightarrow 0)(5-(x^2+2xh+h^2)-5+h^2)/(h)`

`f'(x)=lim_(h\rightarrow 0)(5-x^2-2xh-h^2-5+h^2)/(h)`

`f'(x)=lim_(h\rightarrow 0)(-x^2-2xh)/(h)`

`f'(x)=lim_(h\rightarrow 0)(-x^2)/(h)-(2xh)/(h)`

`f'(x)=lim_(h\rightarrow 0)(-x^2)/(h)-2x`

Apply limit.

`f'(x)=(-x^2)/(0)-2x`

`f'(x)=0-2x`

`f'(x)=-2x`

Therefore, the derivative of given function is -2x.

Answer 2

The derivative of the function f(x) is:

`f'(x)=-2x`

Explanation:

We are given a function f(x) as:

`f(x)=5-x^2`

We have:

`f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)`

( Since,

`(a+b)^2=a^2+b^2+2ab` )

Hence, we get:

`f(x+h)=5-x^2-h^2-2xh`

Also, by using the definition of f'(x) i.e.

`f'(x)= \lim_(h \to 0) (f(x+h)-f(x))/(h)`

Hence, on putting the value in the formula:

`f'(x)= \lim_(h \to 0) (5-x^2-h^2-2xh-(5-x^2))/(h)\\\\\\f'(x)=\lim_(h \to 0) (5-x^2-h^2-2xh-5+x^2)/(h)\\\\i.e.\\\\f'(x)=\lim_(h \to 0) (-h^2-2xh)/(h)\\\\f'(x)=\lim_(h \to 0) (-h^2)/(h)+(-2xh)/(h)\\\\f'(x)=\lim_(h \to 0) -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x`

Hence, the derivative of the function f(x) is:

`f'(x)=-2x`